Question with imaginary number?

Question

Solve (x + iy)^2 = x + iy
Given that x and y are real numbers.
Thanks =)

Answer 1

let x+iy = z

then z^2 = z
or z(z-1) = 0
z = 0 => x,y both zero
or z=1 => x= 1 and y = 0

Answer 2

You could do this by multiplying out the bracket and then equating real and imaginary parts. This would give you a pair of simultaneous equations to solve.
Far easier is to replace x + iy by z to give you z^2 = z.

Answer 3

(x+iy)^2 = x+iy

x^2 + 2ixy - y^2 = x + iy
x^2 - y^2 - x +i(2xy - y) = 0

So....

x^2 - y^2 - x = 0 and
y(2x-1) = 0

Second equation tells us that either y=0,
in which case:
x(x-1)=0, so x=0,1

or x=1/2 in which case:
1/4 - 1/2 = y^2, but y has been defined as real, so
only solution is:

x=0,1
y=0